Zcos
Score Card
Quantitative Aptitude
Total Ques
Attempt
Correct
Score
Easy
24 Ques
0
0
24
Medium
196 Ques
0
2
194
Hard
30 Ques
0
0
30
General Instruction:
Test contains total 20 questions.
Time allowed to finish is 30 mins.
There is no negative marking.
Question - 1.
1. 2X² - 7X + 6 = 0
2. 4Y² + 11Y + 7 = 0
X < Y
X > Y
X ≤ Y
X ≥ Y
Solution:
2X² - 7X + 6 = 0
2X² - 4X - 3X + 6 = 0
2X(X - 2) - 3(X - 2) = 0
(X - 2)(2X - 3) = 0
X = 2 , -3/2
4Y² + 11Y + 7 = 0
4Y² + 7Y + 4Y + 7 = 0
Y(4Y + 7) + 1(4Y + 7) = 0
(Y + 1)(4Y + 7) = 0
Y = -1, -7/4
Question - 2.
1. X - √81 = 0
2. Y² - 81 = 0
X < Y
X > Y
X ≤ Y
X ≥ Y
Solution:
X - √81 = 0
X = 9
Y² - 81 = 0
Y = +9 or -9
Question - 3.
What approximate value should come in place of the question mark in the following questions?
√(8939) x (6.004)² = ?
3050
3390
3200
3350
Solution:
√(8939) x (6.004)² = ?
94 x 36 = 3384
Question - 4.
What approximate value should come in place of the question mark in the following questions?
(323/55) x (971/251) x (56/61) = ?
25
26
23
21
Solution:
(323/55) x (971/251) x (56/61) = ?
(323 x 971 x 56)/(55 x 251 x 61) = 17563448/842105 = 20.85 = 21
Question - 5.
What approximate value should come after solve questions.
32.1 x 2799 ÷ 549 + 110
284
290
295
321
Solution:
32.1 x 2799 ÷ 549 + 110
= 33 x (2800/550) + 110
= 168 + 110 = 278
Question - 6.
Seena, Reena and Ravi begin to jog around a circular stadium and they complete their revolutions in 54 seconds, 42 seconds and 63 seconds respectively. After approximately how many minutes will they come together at the starting point ?
5 minutes
6 minutes
7 minutes
8 minutes
Solution:
L.C.M of 54sec, 42sec and 63sec = 378 sec/60 = 6.3 min = 7 min
Question - 7.
Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.?
10 kmph
11 kmph
12 kmph
13 kmph
Solution:
Let the distance travelled by X km.
Then, (X/10) - (X/15) = 2
3X - 2X = 60
X = 60
Time taken to travel 60 km at 10 km/hr = (60/10)hrs = 6 hrs.
So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.
Required speed = (60/5) kmph = 12 kmph
Question - 8.
Milan rides her bike at an average speed of 30 km/hr and reaches her destination in 6 hours. Heena covers the same distance in 4 hours. If Milan increases her average speed by 10 km/hr and Heena increases her average speed by 5 km/hr, what would be the difference in their time taken to reach the destination ?
54 min.
58 min.
60 min.
64 min.
Solution:
Total distance = 30 x 6 = 180 km
Milan speed = 30 kmph
Reena speed = 45 kmph
New speed of Milan = 40 kmph
New speed of Reena = 50 kmph
Milan time taken to cover distance = 180/40 = 2.5 hr
Reena time taken to cover distance = 180/50 = 3.6 hr
Difference between to time taken = 4.5 - 3.6 = 0.9 x 60 min = 54 min.
Question - 9.
in How many different ways can the letters of the word ORANGE be arranged ?
120
720
512
640
Solution:
Arrange letters of word ORANGE = 6 x 5 x 4 x 3 x 2 x1 = 720
Question - 10.
The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that number?
3
4
5
6
Solution:
Let the ten's digit be x and unit's digit be y.
Then, (10X + Y) - (10Y + X) = 36
9(X - Y) = 36
X - Y = 4
Question - 11.
Out of the total number of students in a college 12% are interested in sports. (3/4) of the total number of students are interested in dancing. 10% of the total number of students are interested in singing and the remaining 15 students are not interested in any of the activities. What is the total number of students in the college ?
450
500
550
600
Solution:
12% students are intrested in sports.
3/4 or 75% students are intrested in dancing and 10% students are intrested in singing.
15 students are not intrested in any activities.
Total percentage = 100% - (12% + 75% + 10%)
Total percentage = 3%
3% of Total students = 15
so, Total students = 500
Question - 12.
The ratio of the present ages of Anu and her mother is 1:2. The ratio of their ages 6 years hence would be 11:20. What was the ratio of their ages 9 years ago ?
5:2
5:11
2:7
1:4
Solution:
Present age of Anu and Her mother X and 2X.
Ages of 6 years hence Anu and Her mother is 11 and 20.
(X + 6) : (2X + 6) 11: 20
X= Anu age = 54
2X Anu mother age = 108
9 years ago Anu and her mother age = 45 : 99 = 5 : 11
Question - 13.
Ravi invested a certain amount at the rate of 6 p.c.p.a. and obtained a simple interest of Rs. 8,730 at the end of 3 years. What amount of compound interest would he obtain on the same amount at the same of interest at the end of 2 years ?
Rs 5,820
Rs 5,949.60
Rs 5,900
Rs 5,994.60
Solution:
Simple interest = 8730
Rate of interest = 6 p.c.p.a.
Time = 3 years
Principal = (S.I. x 100)/(Rate of interest x time)
P = (8730 x 100)/(6 x 3) = 48,500
Then, Compound interest = P[1 + (R/100)]² - P
C.I. = 48,500[1 + (6/100)]² - 48500
C.I. = 48500 x (53/50) x (53/50) - 48500
C.I. = 54,494.6 - 48500
C.I. = 5,994.6
Question - 14.
Car A travels at the speed of 65 km/hr and reaches its destination in 8 hours. Car B travels at the speed of 70 km/hr and reaches its destination in 4 hours. What is the ratio of the distance covered by car A and car B respectively ?
12 : 5
13 : 7
15 : 7
11 : 7
Solution:
Distance travelled by Car A = 65 kmph x 8 hr = 520 km
Distance travelled by Car B = 70 kmph x 4 hr = 280 km
Ratio between the distance = 520 : 280
Ratio between the distance = 13 : 7
Question - 15.
The ratio of the students in schools A, B and C is 5 : 4 : 7. If the number of students in the schools are increased by 20%, 25% and 20% respectively, what would be the new ratio of the students in schools A, B and C ?
5 : 5 : 7
25 : 30 :42
30 : 25 : 42
42 : 20 : 25
Solution:
Lets Students of School A, B and C respectivily 50, 40 and 70.
Then Ratio = 5 : 4 : 7
Increased students in school A = 50 + (50 x 20/100) = 60
Increased students in school B = 40 + (40 x 25/100) = 50
Increased students in school C = 70 + (70 x 20/100) = 84
Then New ratio = 60 : 50 : 84
Ratio = 30 : 25 : 42
Question - 16.
What approximate value should come in place of the question mark in the following questions?
74% of 366 + 12.6% of 317 = ?
315
310
320
305
Solution:
74% of 366 + 12.6% of 317 = ?
= (74/100) x 366 + (12.6/100) x 317
= (74 x 366)/100 + (12.6 x 317)/100
= 27,084/100 + 3994.2/100
= 310.782
310
Question - 17.
The angles of the quadrilateral are in the ratio of 2 : 4 : 7 : 5. The smallest angle of the quadrilateral is equal to the smallest angle of a triangle. One of the angle of the triangle is twice the smallest angle of the triangle. What is the second largest angle of the triangle ?
80
70
60
50
Solution:
The angles of the quadrilateral are in the ratio of 2 : 4 : 7 : 5
Total angle of quadrilateral are = 2X + 4X + 7X + 5X = 18X
18X = 360
X = 20 degree
Smallest angle of qudrilateral = 2X = 2 x 20 = 40 degree
Smallest angle of triangle = 40
Angle is twice of smallest angle = 2 x 40 = 80 degree
Third angle of Triangle = 180 - (40 + 80) = 60
Second largest angle = 60 degree
Question - 18.
Two men alone or three women alone can complete a piece of work in 4 days. In how many days can 1 woman and one man together complete the same piece of work ?
6 days
24/5 days
12/1.75 days
6 days
Solution:
2 men = 2 women
1 man + 1 woman = [(3/2) + 1] women = 5/2 women
M1D1 = M2D2
3 x 4 = (5/2) x D2
D2 = (3 x 4 x 2)/5 = 24/5 days
Question - 19.
A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?
240 m
260 m
120 m
320 m
Solution:
Crossing time = (length of train + Length of platform)/Relative speed
20seconds = length of train/[(54 x 5)/18]
20 seconds = L/15m/sec
L = 300 metre
Again,
Crossing time = (length of train + Length of platform)/Relative speed
36 sec = (300 + Lenght of platform)/15 m/sec
300 + Lenght of platform = 540 metre
Lenght of platform = 540 - 300
Lenght of platform = 240 metre
Question - 20.
A alone can do a piece of work in 6 days and B alone in 8 days. A and B undertook to do it for Rs. 3200. With the help of C, they completed the work in 3 days. How much is to be paid to C?
Rs 450
Rs 500
RS 525
Rs 400
Solution:
A, one day work = 1/6
B, one day work = 1/8
Part of work done by C = 1 - (1/6 + 1/8)
Part of work done by C = 1/8
Total money is 3200.
So, part of C = 3200/8 = 400
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